Question

In a Young's double-slit experiment, two light waves, each of intensity \( I_0 \), interfere at a point, having a path difference \( \frac{\lambda}{8} \) on the screen. Find the intensity at this point.

Hint:

In interference patterns, the intensity is determined by the phase difference, which depends on the path difference between the waves.

Answer 1

The intensity in an interference pattern is given by: \[ I = I_0 \left( 1 + \cos \delta \right) \] where \( \delta \) is the phase difference given by: \[ \delta = \frac{2\pi}{\lambda} \cdot \text{path difference} \] Substitute the given path difference \( \frac{\lambda}{8} \): \[ \delta = \frac{2\pi}{\lambda} \cdot \frac{\lambda}{8} = \frac{\pi}{4} \] Thus, the intensity is: \[ I = I_0 \left( 1 + \cos \frac{\pi}{4} \right) = I_0 \left( 1 + \frac{\sqrt{2}}{2} \right) \] \[ I = I_0 \left( 1 + 0.707 \right) = 1.707 I_0 \] Thus, the intensity at the point is \( 1.707 I_0 \). 

Answer 2

In a Young's double-slit experiment, two light waves with intensity \( I_0 \) interfere at a point on the screen, and the path difference between them is \( \frac{\lambda}{8} \). We are tasked with finding the intensity at this point. 

1. Understanding the Concepts: 

- Interference of Light: In an interference experiment, such as Young's double-slit, light waves from two slits interfere with each other. The resulting intensity depends on the phase difference between the waves at the point where they meet.

- Path Difference and Phase Difference: The phase difference (\( \Delta \phi \)) between the two waves is related to the path difference (\( \Delta x \)) by the equation:

\[ \Delta \phi = \frac{2 \pi \Delta x}{\lambda} \]

- Here, the path difference is \( \Delta x = \frac{\lambda}{8} \), so the phase difference is:

\[ \Delta \phi = \frac{2 \pi \times \frac{\lambda}{8}}{\lambda} = \frac{\pi}{4} \]

2. Intensity in Interference:

The total intensity \( I \) at the point where two waves interfere is given by the formula:

\[ I = I_1 + I_2 + 2 \sqrt{I_1 I_2} \cos(\Delta \phi) \]

- Since both waves have the same intensity \( I_0 \), the formula becomes:

\[ I = I_0 + I_0 + 2 \sqrt{I_0 I_0} \cos(\Delta \phi) \]

\[ I = 2I_0 + 2I_0 \cos(\frac{\pi}{4}) \]

3. Calculating the Intensity:

We know that \( \cos\left(\frac{\pi}{4}\right) = \frac{1}{\sqrt{2}} \), so the intensity is:

\[ I = 2I_0 + 2I_0 \times \frac{1}{\sqrt{2}} \]

\[ I = 2I_0 \left( 1 + \frac{1}{\sqrt{2}} \right) \]

Now, simplifying this expression:

\[ I = 2I_0 \left( \frac{\sqrt{2} + 1}{\sqrt{2}} \right) \]

4. Final Answer:

The intensity at the point on the screen where the path difference is \( \frac{\lambda}{8} \) is:

\[ I = 2I_0 \left( \frac{\sqrt{2} + 1}{\sqrt{2}} \right) \]